Why there needs to be a $ in calls like “runSomeMonad $ do …”?

Question

Apparently the only possible interpretation of runSomeMonad do ... is runSomeMonad (do ...). Why isn\'t the first variant allowed by the Haskell sy

Answer 1

Note that you can observe this effect with not just do, but also let, if, \, case, the extensions mdo and proc…and the dread unary -. I cannot think of a case in which this is ambiguous except for unary -. Here’s how the grammar is defined in the Haskell 2010 Language Report, §3: Expressions.

exp
    → infixexp :: [context =>] type
    | infixexp

infixexp
    → lexp qop infixexp
    | - infixexp
    | lexp

lexp
    → \ apat1 … apatn -> exp
    | let decls in exp
    | if exp [;] then exp [;] else exp
    | case exp of { alts }
    | do { stmts }
    | fexp

fexp
    → [fexp] aexp

aexp
    → ( exp )
    | …

There just happens to be no case defined in fexp (function application) or aexp (literal expression) that allows an unparenthesised lexp (lambda, let, etc.). I would consider this a bug in the grammar.

Fixing this would also remove the need for the $ typing hack.